将共享指针存储矢量对象抛出错误？

Question

喜IM存储一个矢量对象共享指针和调用打印过程打印出来，但它抛出一个错误，因为“敌不过呼叫**”下面是代码

namespace info 
{ 
typedef struct { 
     std::string Yes; 
     std::string NO; 
     int ID_NO; 
} mystruct; 

typedef std::vector<mystruct> myvector; 

typedef struct { 
     bool status; 
     std::shared_ptr<myvector> shared; 
} Myreturn; 

} 

void printmyinfo(const std::shared_ptr<info::mystruct>& printval) 
{ 
     if(printval == NULL) 
     { 
       cout<<"Sorry no value exist"<<endl; 
       return; 
     } 
     // here is print code which is good. 
} 


int main() 
{ 
     info::mystruct *structinfo=new mystruct ; 
     structinfo->yes="okay"; 
     structinfo->ID_NO=10; 
     info::myvector vectorobject; 
     vectorobject.push_back(structinfo); 
     info::Myreturn *returnobj=new Myreturn; 
     returnobj->shared(myvector)// bad how to store vector in share pointer of Myreturn structure 

//如何用myvector填充Myreturn.shared并传递给printmyinfo？ printmyinfo（myvector）//坏 }

在这一行returnobj.shared(structinfo)错误thow请帮我解决它感谢

如何填写Myreturn.shared与myvector，并传递给printmyinfo？

Answer 1

也许你想要这个。

returnobj.shared = shared_ptr<info::myvector>(&structinfo) // 

还纠正最后一行，您的printmyinfo()参数类型不匹配。